SOLUTION: A cannonball fired out to sea from a shore battery follows a parabolic trajectory given by the graph of the equation h(x)=10x - 0.01x^2 where h(x) is the height of the cannonball

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Question 169462: A cannonball fired out to sea from a shore battery follows a parabolic trajectory given by the graph of the equation
h(x)=10x - 0.01x^2
where h(x) is the height of the cannonball above the water when it has traveled a horizontal distance of x feet.
a.) What is the maximum height that the cannonball reaches?
b.) How far does the cannonball travel horizontally before splashing into the water?

Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
since this is a parabola in the form of ax^2 + bx + c, we can find the x value of the maximum / minimum point by using the formula -b/2a, where b is the coefficient of the x term and a is the coefficient of the x^2 term.
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a = -.01
b = 10
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since a is negative, the graph will be head up and tails down so that the maximum / minimum point will be a maximum point.
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-b/2a = -(10)/2*(-.01) = (-10)/(-.02) = 10/.02 = 500
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x = 500 is the x value of the maximum point.
h(500) will be the y value.
since h(x) = 10x - .01x^2, then
h(500) = 10*500 - .01*(500)^2
simplifying:
h(500) = 5000 - .01*250000
h(500) = 5000 - 2500
h(500) = 2500
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maximum height will be 2500 feet when the cannon ball has reached a horizontal distance of 500 feet.
graph of this trajectory looks like this:
look below the graph for further comments.

the x intercepts are when the graph of the equation crosses the x axis.
these are determined by setting the quadratic equation equal to 0.
10x - .01x^2 = 0
this can be factored to become:
x (10-.01x) = 0
x = 0
or
10-.01x = 0
solving for x in equation:
10 - .01x = 0
add .01x to both sides of equation:
10 = .01x
divide both sides of equation by .01:
10/.01 = x
simplify:
1000 = x
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x intercepts are either
x = 0
or
x = 1000
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this can be seen from the graph.
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